Question

How do you find the equation of the line tangent to `y=secx`, at (pi/3,2)?

Answer 1

`y=2sqrt3x-(2pisqrt3)/3+2`

Explanation:

Name the point `" "(pi/3,2)` by A.
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The line passes through the point `A(pi/3,2)" "`.
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The slope of this tangentbis determined by performing `y'` at
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`x=pi/3`
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Differentiating `y`
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`y' = (secx)'=tanxsecx`
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The slope of tangent line at `x=pi/3` is:
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`y'_(x=pi/3) =tan(pi/3)sec (pi/3)=sqrt3 xx 2=color (brown)(2sqrt3)`
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The equation of the tangent line with slope`" "color (brown)(2sqrt3)`
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passing through`" "A (pi/3,2)` is:
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`y-y_A= y_(x_A)(x-x_A)`
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`y-2=2sqrt3 (x-pi/3)`
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`y=2sqrt3x-(2pisqrt3)/3+2`